h=(E-5)/15

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Solution for h=(E-5)/15 equation: 



h=(-5)/15

We move all terms to the left:

h-((-5)/15)=0

We multiply all the terms by the denominator


h*15)-((-5)=0

Wy multiply elements

15h^2=0

a = 15; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·15·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$h=\frac{-b}{2a}=\frac{0}{30}=0$

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